Answer:
-600 J
Explanation:
F₁ = 8i +29 j + 32k
F₂ = 48 i - 59 j - 22 k
F = F₁ +F₂ = 8i +29 j + 32k +48 i - 59 j - 22 k
F = 56i - 30 j + 10 k
displacement d = ( 0 - 20 )i + ( 0 - 15 )j + ( 7 -0) k
d = - 20 i - 15 j + 7 k
Work Done = F dot product d
F . d = - 56 x 20 - 30 x - 15 + 10 x 7
= - 1120 +450 + 70
= -600 J
Answer:
12,608 kPa
Explanation:
First we need to convert the density to standard units, that is kg/m^3
0.85 g/cm^3 = 0.85 ×
= 0.85 × 1000 = 850 kg/m^3
Pressure of oil = pressure at surface + rho × g × h
= 101,000 + (850×9.81×1500)
= 12,608 kPa
The units in the part 'rho.g.h' will cancel out against each other and you will be left with the unit Pascals - which is the unit for pressure.
Hope that answers the question, have a great day!
Answer:
ºC
Explanation:
First, let's write the energy balance over the duct:

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

So, let's isolate
:

The Cp of the air at 27ºC is 1007
(Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are
and Q.
Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.
The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:
Perimeter:

Surface area:

Then, the heat Q is:

Finally, find the exit temperature:

=27.0000077 ºC
The temperature change so little because:
- The mass flow is so big compared to the heat flux.
- The transfer area is so little, a bigger length would be required.
R1 + R4 = 1430 + 1350 = 2780 = R14 series combination of R1 & R4
R2 + R5 = 1350 + 1150 = 2500 = R25
The circuit has been reduced to 3 resistors in parallel
R314 = 2780 * 1100 / (2780 + 1100) = 788 this is the resistance of the parallel combination of R14 and R3
R31425 = 2500 * 788 / (2500 + 788) = 599 which is the equivalent of the circuit - you can also use the formula for 3 resistors in parallel but this seems simpler
That ratio is 2 .
<h3>What is ratio?</h3>
A ratio is the comparison of the two numbers bydivision.
Taking the first two outputs, or the -1/8 and -1/4, we can divide not the second one by the first one to find the
ratio:
-1/4-1/8
When dividing fractions, we multiply by the reciprocal:
-1/4x-8/1
To multiply fractions, multiply straight across:
(-1x-8)/(4x1) 8/4=2
To know more about ratio click-
brainly.com/question/25927869
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